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Dear community,

I'm looking for a table that lists the full commutation relations for the Chevalley group $G_2$. I could do this myself (although it would take quite a while), but maybe there is already such a table out there?

Sincerely, and thank you,

Moshe Adrian

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  • $\begingroup$ Jut to be sure: you want the commutation relations for the group or the Lie algebra? $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 10, 2012 at 1:49
  • $\begingroup$ Dear Mariano : Yes, I'd like the commutation relations for the group. $\endgroup$
    – Moshe Adrian
    Apr 10, 2012 at 2:22
  • $\begingroup$ Would the relations between 6 or 7 consecutive root groups of $G_2(q)$ be sufficient for your purposes? I can give you that if you want. $\endgroup$
    – Koen S
    Apr 10, 2012 at 11:18
  • $\begingroup$ @ Koen : thank you for your response, Jim has now provided me with what I need. $\endgroup$
    – Moshe Adrian
    Apr 10, 2012 at 18:05
  • $\begingroup$ see page 443 of jstor.org/stable/2372888?seq=1#page_scan_tab_contents $\endgroup$
    – Q. Zhang
    Aug 19, 2016 at 17:01

3 Answers 3

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It would help to clarify the context of "Chevalley group" here, but over any field one gets uniform results by starting with a Chevalley basis for the associated complex Lie algebra. The commutation relations for rank 2 groups including $G_2$ are given explicitly in SGA3 and written down similarly in my 1975 Springer text Linear Algebraic Groups in section 33.5.

ADDED: There are two cases not involving a pair of positive roots. 1) If the two roots are independent, they both lie in some positive system. Using the standard picture of the 12 roots, it's easy to visualize the Weyl group transformation involved. Then just relabel the two roots by the corresponding two in the fixed positive system and apply the formulas worked out there. 2) If the roots are negatives of each other, life gets a little more complicated. But here the commutation takes place in a rank 1 group. Explicit commutation formulas here are usually avoided in the literature but take place just in a group like $\rm{SL}_2$.

Maybe it would also help to add two relevant online references. Both are based ultimately on a Chevalley basis (keeping in mind that such a basis is unique only up to certain sign choices) as well as the ideas in Chevalley's 1956-58 classification seminar.

SGA3 is now online here. See in particular Expose XXIII.

Steinberg's 1967-68 Yale lectures on Chevalley groups are available at his homepage here. See Section 6, page 66.

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  • $\begingroup$ Dear Jim, This is exactly what I am looking for, however : it seems that in the references you mentioned, there are only formulas for $[\epsilon_{\delta}(y), \epsilon_{\gamma}(x)]$ where $\delta, \gamma$ are positive roots. Is there a reference where $\delta, \gamma$ can be arbitrary roots? (many commutators will be trivial, but it is important for me to allow $\delta$ or $\gamma$ to be negative) Moshe $\endgroup$
    – Moshe Adrian
    Apr 10, 2012 at 15:01
  • $\begingroup$ @Moshe: Maybe my added comments will help to clarify things. $\endgroup$
    – Jim Humphreys
    Apr 10, 2012 at 17:18
  • $\begingroup$ Concerning Steinberg's Yale lectures, his homepage at UCLA was eventually removed after his death, though the notes might resurface in published form someday. Meanwhile, Chris Drupieski has produced a LaTeX version: math.depaul.edu/cdrupies/research/papers/chevalleygroups.pdf Also, the 1956-58 Chevalley seminar is available online at Numdam: numdam.org/numdam-bin/feuilleter?id=SCC_1956-1958__2_ $\endgroup$
    – Jim Humphreys
    Sep 10, 2016 at 13:42
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At first: sorry, I could't find the source you were looking for, but I'm so free to sketch two ways how you in my opinion get very fast to the informations you need (especially the latter, generic, might be the reason rarely somebody wants to work out all relations?)

Maybe you've a more specific question than just knowing the entire table, which schould be quite huge?

By the way: With Chevalley-group you mean the finite simple group? But not the twisted one? I assume yes on both, as this is what you wrote ;-)

1) The probably easiest way to directly and elementary get ahold of $G_2$ is as folding of $D_4$ by the cyclic interchange of the three outer nodes. I.e. take $D_4=so_8$ and take the sub-algebra, sub-root-system, etc. fixed by the outer order-3-automorphism often named triality. I'm not 100% sure and don't have a proper source, where this is done for Chevalley groups, but I'm pretty positive, you just do it ;-)

....do NOT confuse this with the construction of the Steinberg group $^3D_4(q^3)$, where one takes (analogously to the construction of unitary groups) the elements, where the action of the automorphisms matches the order 3 galois field automorphisms forced to exist by taking $\mathbb{F}_{q^3}$.

This way you're commutator relations of $G_2(q)$ are just a subset of those in $SO_8(\mathbb{F}_q)$, where calculating is obvious (but tedious?)

2) The more uniform, direct way would be starting with a Chevalley basis of the Lie algebra $G_2$ (which I'm sure is not hard and/or can be found in literature - I'll look into it, if I find time...) - the relations are then just exponentiated to the Chevalley group.

This way you can read off the commutator relations just from the structure constants of the Lie algebra $G_2$, that I'm sure are worked out at many places

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  • $\begingroup$ @ Simon : Thank you for your response, 2) is the approach I was hoping to take, and Jim has kindly provided a resolution of this. $\endgroup$
    – Moshe Adrian
    Apr 10, 2012 at 19:25
  • $\begingroup$ Yes, that was definitely the more precise and helpful answer - but thanks for the thanks at this place too ;-) ;-) I looked at his Lie algebra book but only found the "reader problem", whether the usual G2 basis is a Chevalley one... $\endgroup$
    – Simon Lentner
    Apr 10, 2012 at 23:42
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I once calculated the commutators, based on the book Notes on Lie Algebras, by Samelson.

Suppose $\alpha_1$ and $\alpha_2$ are the simple roots, with $\alpha_1$ the long root and $\alpha_2$ the short root.

Thus all positive roots are \begin{equation*} \alpha_1, \alpha_2, \alpha_1 + \alpha_2, \alpha_1 + 2 \alpha_2, \alpha_1 + 3 \alpha_2, 2 \alpha_1 + 3\alpha_2 \end{equation*}

Here is a list of all commutators of $G_2$. Because the list is long, I organize it as follows.

Commutators of the form $[h , e ] $
If $\alpha = r \alpha_1 + s \alpha_2$ then $[ h_{\alpha_1} , e_{\alpha} ] = (2 r -s ) e_{\alpha}$ and $[ h_{\alpha_2} , e_{\alpha} ] = (-3 r + 2 s ) e_{\alpha}$

Commutators of the form $[ e_{\alpha}, e_{-\alpha} ] $
Because of the antisymmetry of the commutator, I only write down the commutators for $\alpha \ge 0$ \begin{align*} [e_{\alpha_2},e_{-\alpha_2}] & = h_{\alpha_2} & [e_{\alpha_1},e_{-\alpha_1}] & = h_{\alpha_1} \\ [e_{\alpha_1+\alpha_2},e_{-\alpha_1-\alpha_2}] & = 3 h_{\alpha_1}+h_{\alpha_2}& [e_{\alpha_1+2 \alpha_2},e_{-\alpha_1-2 \alpha_2}] & = 3 h_{\alpha_1}+2 h_{\alpha_2}\\ [e_{\alpha_1+3 \alpha_2},e_{-\alpha_1-3 \alpha_2}] & = h_{\alpha_1}+h_{\alpha_2}& [e_{2 \alpha_1+3 \alpha_2},e_{-2 \alpha_1-3 \alpha_2}] & = 2 h_{\alpha_1}+h_{\alpha_2} \end{align*}
Commutators of the form $[ e_{\alpha}, e_{\beta} ] $ with $\alpha + \beta$ a root and $\alpha + \beta \neq 0$
Because of the antisymmetry of the commutator, I only write down the commutators for $\alpha \le \beta$ \begin{align*} [e_{-\alpha_2},e_{-\alpha_1}] & = -e_{-\alpha_1-\alpha_2} & [e_{-\alpha_1-2 \alpha_2},e_{\alpha_1+\alpha_2}] & = -2 e_{-\alpha_2} \\ [e_{-\alpha_2},e_{-\alpha_1-\alpha_2}] & = -2 e_{-\alpha_1-2 \alpha_2} & [e_{-\alpha_1-2 \alpha_2},e_{\alpha_1+3 \alpha_2}] & = -e_{\alpha_2} \\ [e_{-\alpha_2},e_{-\alpha_1-2 \alpha_2}] & = -3 e_{-\alpha_1-3 \alpha_2} & [e_{-\alpha_1-2 \alpha_2},e_{2 \alpha_1+3 \alpha_2}] & = e_{\alpha_1+\alpha_2} \\ [e_{-\alpha_2},e_{\alpha_1+\alpha_2}] & = 3 e_{\alpha_1} & [e_{-\alpha_1-3 \alpha_2},e_{\alpha_2}] & = e_{-\alpha_1-2 \alpha_2} \\ [e_{-\alpha_2},e_{\alpha_1+2 \alpha_2}] & = 2 e_{\alpha_1+\alpha_2} & [e_{-\alpha_1-3 \alpha_2},e_{\alpha_1+2 \alpha_2}] & = -e_{-\alpha_2} \\ [e_{-\alpha_2},e_{\alpha_1+3 \alpha_2}] & = e_{\alpha_1+2 \alpha_2} & [e_{-\alpha_1-3 \alpha_2},e_{2 \alpha_1+3 \alpha_2}] & = -e_{\alpha_1} \\ [e_{-\alpha_1},e_{-\alpha_1-3 \alpha_2}] & = -e_{-2 \alpha_1-3 \alpha_2} & [e_{-2 \alpha_1-3 \alpha_2},e_{\alpha_1}] & = e_{-\alpha_1-3 \alpha_2} \\ [e_{-\alpha_1},e_{\alpha_1+\alpha_2}] & = -e_{\alpha_2} & [e_{-2 \alpha_1-3 \alpha_2},e_{\alpha_1+\alpha_2}] & = -e_{-\alpha_1-2 \alpha_2} \\ [e_{-\alpha_1},e_{2 \alpha_1+3 \alpha_2}] & = e_{\alpha_1+3 \alpha_2} & [e_{-2 \alpha_1-3 \alpha_2},e_{\alpha_1+2 \alpha_2}] & = e_{-\alpha_1-\alpha_2} \\ [e_{-\alpha_1-\alpha_2},e_{-\alpha_1-2 \alpha_2}] & = 3 e_{-2 \alpha_1-3 \alpha_2} & [e_{-2 \alpha_1-3 \alpha_2},e_{\alpha_1+3 \alpha_2}] & = -e_{-\alpha_1} \\ [e_{-\alpha_1-\alpha_2},e_{\alpha_2}] & = 3 e_{-\alpha_1} & [e_{\alpha_2},e_{\alpha_1}] & = e_{\alpha_1+\alpha_2} \\ [e_{-\alpha_1-\alpha_2},e_{\alpha_1}] & = -e_{-\alpha_2} & [e_{\alpha_2},e_{\alpha_1+\alpha_2}] & = 2 e_{\alpha_1+2 \alpha_2} \\ [e_{-\alpha_1-\alpha_2},e_{\alpha_1+2 \alpha_2}] & = -2 e_{\alpha_2} & [e_{\alpha_2},e_{\alpha_1+2 \alpha_2}] & = 3 e_{\alpha_1+3 \alpha_2} \\ [e_{-\alpha_1-\alpha_2},e_{2 \alpha_1+3 \alpha_2}] & = -e_{\alpha_1+2 \alpha_2} & [e_{\alpha_1},e_{\alpha_1+3 \alpha_2}] & = e_{2 \alpha_1+3 \alpha_2} \\ [e_{-\alpha_1-2 \alpha_2},e_{\alpha_2}] & = 2 e_{-\alpha_1-\alpha_2} & [e_{\alpha_1+\alpha_2},e_{\alpha_1+2 \alpha_2}] & = -3 e_{2 \alpha_1+3 \alpha_2} \\ \end{align*}



A comment about the signs
After much trial and error I found that the signs in the commutators above are given by the following quite compact expression

1) Define a function $F: \text{Roots} \to \{0,1\}$
\begin{equation*} F(\alpha) = 0 \quad\text{if}\quad \alpha \in \big\{(0,1),(2,3),(0,-1),(-2,-3)\big\} \quad\text{and}\quad F(\alpha) = 1 \quad\text{otherwise} \end{equation*}
2) Define a total order $\prec$ as
\begin{align*} - \alpha_1 - 2 \alpha_2 \prec - \alpha_1 \prec 2 \alpha_1 + & 3 \alpha_2 \prec - \alpha_1 - 3 \alpha_2 \prec - \alpha_1 - \alpha_2 \prec - \alpha_2 \prec \\ & \prec \alpha_2 \prec \alpha_1 + \alpha_2 \prec \alpha_1 +3 \alpha_2 \prec -2 \alpha_1 - 3 \alpha_2 \prec \alpha_1\prec \alpha_1 + 2 \alpha_2 \end{align*}
3) The sign in the commutator $[e_{\alpha} , e_{\beta} ] = n_{\alpha \beta}e_{\alpha+\beta}$ is then given by
\begin{equation*} (-1)^{F(\alpha) F(\beta)} \quad\text{if}\quad \alpha\prec\beta \end{equation*} and \begin{equation*} - (-1)^{F(\alpha) F(\beta)} \quad\text{if}\quad \alpha\succ\beta \end{equation*}

I wrote down some details in a blog post.

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